Proving the periodicity of function I just wanted someone to verify the proof i have written

Problem statement: If $f(x)$ is periodic with period $T$, then $f(ax+b)$ is  periodic with period $T/a$, $a>0$

My procedure:
1) $f(x) = f(x+T)$; (since $f(x)$ is periodic with period $T$)
2) Replacing $x$ by $ax+b$, $f(ax+b)$= $f(ax+b+T)$(since f(x) is periodic with T and this means $T$ being the smallest no. for the repetition of value $f(x)$ ) = $f(a[x+T/a]+b)$ 
3) 2) shows that $x+T/a$ is the point of repetition of value $f(ax+b)$ for any $x$ belonging to domain  and could be the period of $f(ax+b)$,for $T/a$ really to be the period , it has to be the smallest no. for repetition of $f(ax+b)$
4) Let us assume that a period for $f(ax+b)$ exists $T_1$(say) and $**T_1$<$(T/a)$, then $f(a(x+T_1)+b)$=$f(ax+b)$=$f(ax+aT_1+b)$ (we assumed that $T_1$ is the period)**
5)from 2)  $f(ax+b)$= $f(ax+b+T)$ = $f(ax+aT_1+b)$(from 4) )= $f(ax+(aT_1/T)T+b)$, but  $f(ax+b)= f(ax+b+T)$ shows that $T$ is the period of $f(x)$(see prob. statement and 1) ) thus in $f(ax+(aT_1/T)T+b)$ , since $T$ is the period of $f(x)$, $(aT_1/T) \geq 1$ or $T1 \geq (T/a)$, thus contradicting our assumption $T_1$ as a period and thus $T/a$ is the period of $f(ax+b)$.
Did I do correctly?
 A: I don't follow your step 2 at the end. Maybe you need more parentheses? Why does that equal the previous expression?

If $g(x)=f(x+b)$, the graph of $g$ is the graph of $f$ shifted left by $b$ units. So $g$ has the same period as $f$.
If $h(x)=g(ax)$, the graph of $h$ is the graph of $g$ scaled horizontally toward/away from the $y$-axis by a factor of $\frac{1}{a}$. So $h$ has a period that is $\left|\frac{1}{a}\right|$ times the period of $g$.
Hence, the period of $h(x)=g(ax)=f(ax+b)$ is $T\cdot\frac{1}{|a|}$, or just $\frac{T}{|a|}$.
A: A much simpler way would be to say that $f(x) = f(x+T)$, so $f(ax+b) = f(ax+b+T) = f(a(x+\frac{T}{a}) + b)$ which means that the function repeats with period $x+\frac{T}{a}$.
However, it looks as if your method is correct, too. It seems like after 2) you are done (if you mean with parenthesis, $f(a(x+\frac{T}{a})+b)$ and the rest seems to be verification that is indeed the smallest period?
A: Let $g(x):=f(ax+b)$.
As$$g(x+\frac Ta)=f(ax+T+b)=f(ax+b)=g(x),$$
the function $g$ has obviously the period $\dfrac Ta$.
Now assume $g$ has a shorter period, let $\dfrac{T'}a$ with $T'<T$. We have
$$f(x+T')=g\left(\frac{x+T'-b}a\right)=g\left(\frac{x-b}a\right)=f(x),$$
so that the period of $f$ is $T'$, a contradiction.
